A (known?) hypergeometric identity Incidentally I've obtained a hypergeometric identity that I've not seen before:
$${}_3F_2(-m,-n,m+n; 1, 1; 1) = \frac{m^2+n^2+mn}{(m+n)^2} {\binom{m+n}{m}}^2$$
So, I wonder if it is well-known and possibly represents a particular case of something more general?
P.S. I've tried to simplify() the l.h.s. in Maple but it did not succeed, giving a hope that the identity is not completely trivial. ;)
EDIT: There seems to be a bug in formula rendering, so I'm repeating it below in plain LaTeX:
{}_3F_2(-m,-n,m+n; 1, 1; 1) = \frac{m^2+n^2+mn}{(m+n)^2} {\binom{m+n}{m}}^2
 A: Your relation is a particular case of the Karlsson--Minton relations (see Section 1.9 in the $q$-Bible by Gasper and Rahman). It's also a contiguous identity to Pfaff--Saalschütz.
EDIT.
First of all I apologise for giving insufficient comments on the problem.
I learned from Max a very nice graph-theoretical interpretation of the identity
which makes good reasons for not burring it in the list of "ordinary" problems.
The hypergeometric series (function)
$$
{}_ {p+1}F_ p\biggl(\begin{matrix} a_ 0,\ a_ 1,\ \dots,\ a_ p \cr
b_ 1,\ \dots ,\ b_ p\end{matrix};x\biggr)
= \sum_ {k=0}^\infty \frac{(a_ 0)_ k(a_ 1)_ k\dots (a_ p)_ k}{(b_ 1)_ k\dots (b_ p)_ k}\frac{x^k}{k!},
$$
where
$$
(a)_ 0=1 \quad\text{and}\quad
(a)_ k =\frac{\Gamma(a+k)}{\Gamma(a)}= a(a+1)\dots (a+k-1) \quad\text{for } k\in \mathbb Z_ {>0}
$$
(I consider the ones with finite domain of convergence $|z|<1$), have very nice
history and links to practically everything in mathematics. There are many
transformation and summation theorems for them, both classical and contemporary.
There are very efficient algorithms and packages for proving them, like
the algorithm of creative telescoping (due to W. Gosper and D. Zeilberger)
and the package HYP
which allows one to manipulate and identify binomial and hypergeometric series
(due to C. Krattenthaler). An example of classical summation theorem is
the Pfaff--Saalschütz sum
$$
{}_ 3F_ 2\biggl(\begin{matrix} -m,\ a,\ b \cr
c,\ 1+a+b-c-m\end{matrix};1\biggr)
=\frac{(c-a)_ m(c-b)_ m}{(c)_ m(c-a-b)_ m}
$$
where $m$ is a negative integer, with a generalisation
$$
{}_ {p+1}F_ p\biggl(\begin{matrix} a,\ b_ 1+m_ 1,\ \dots,\ b_ p+m_ p \cr
b_ 1,\ \dots ,\ b_ p\end{matrix};1\biggr)=0
\quad\text{if } \operatorname{Re}(-a)>m_ 1+\dots+m_ p
$$
and
$$
{}_ {p+1}F_ p\biggl(\begin{matrix} -(m_ 1+\dots+m_ p),\ b_ 1+m_ 1,\ \dots,\ b_ p+m_ p \cr
b_ 1,\ \dots ,\ b_ p\end{matrix};1\biggr)=(-1)^{m_ 1+\dots+m_ p}
\frac{(m_ 1+\dots+m_ p)!}{(b_ 1)_ {m_ 1}\dots (b_ p)_ {m_ p}}
$$
due to B. Minton and Per W. Karlsson (here $m_ 1,\dots,m_ p$ are nonnegative integers).
Max's original identity is not a straightforward particular case but a linear combination
of three contiguous Pfaff--Saalschütz-summable hypergeometric series.
(Two hypergeometric functions are said to be contiguous if they are alike except
for one pair of parameters, and these differ by unity.) Because of having three
hypergeometric functions, I do not see any fun in writing the corresponding details
but indicate a simpler hypergeometric derivation.
Applying Thomae's transformation
$$
{}_ 3F_ 2\biggl(\begin{matrix} -m,\ a,\ b \cr
c,\ d\end{matrix};1\biggr)
=\frac{(d-b)_ m}{(d)_ m}\cdot{}_ 3F_ 2\biggl(\begin{matrix} -m,\ c-a,\ b \cr
c,\ 1+b-d-m\end{matrix};1\biggr)
$$
the problem reduces to evaluation of the series
$$
{}_ 3F_ 2\biggl(\begin{matrix} -m,\ n+1,\ m+n \cr
1,\ n\end{matrix};1\biggr).
$$
Writing
$$
\frac{(n+1)_ k}{(n)_ k}=\frac{n+k}{n}=1+\frac kn
$$
the latter series becomes
$$
{}_ 3F_ 2\biggl(\begin{matrix} -m,\ n+1,\ m+n \cr
1,\ n\end{matrix};1\biggr)
={}_ 2F_ 1\biggl(\begin{matrix} -m,\ m+n \cr 1 \end{matrix};1\biggr)
+\frac{(-m)(m+n)}{n} {}_ 2F_ 1\biggl(\begin{matrix} -m+1,\ m+n+1 \cr 2 \end{matrix};1\biggr)
$$
and the latter two series are summed with the help of the Chu--Vandermonde summation
(a particular case of the Gauss summation theorem).
As for general forms of Max's identity, I can mention that there is no use of the integrality of $n$
in the last paragraph, and I could even expect something a la Minton--Karlsson in general.
